Difference between revisions of "009A Sample Final 2, Problem 1"

Latest revision as of 09:05, 1 December 2017

Compute

(a) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow 4}{\frac {{\sqrt {x+5}}-3}{x-4}}}

(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 0} \frac{\sin^2x}{3x}}

(c) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}}

Solution

Detailed Solution

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@@ Line 6: / Line 6: @@
 <span class="exam">(c) &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}</math>
+<hr>
+[[009A Sample Final 2, Problem 1 Solution|'''<u>Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations: &nbsp;
-|-
-|'''L'Hôpital's Rule, Part 1'''
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow c}f(x)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow c}g(x)=0,</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g</math>&nbsp; are differentiable functions
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;on an open interval &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; containing &nbsp;<math style="vertical-align: -5px">c,</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)\ne 0</math>&nbsp; on &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; except possibly at &nbsp;<math style="vertical-align: 0px">c.</math>&nbsp;
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp; <math style="vertical-align: -18px">\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}.</math>
-|}
+[[009A Sample Final 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''Solution:'''
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We begin by noticing that we plug in &nbsp;<math style="vertical-align: 0px">x=4</math>&nbsp; into
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{\sqrt{x+5}-3}{x-4},</math>
-|-
-|we get &nbsp; <math style="vertical-align: -12px">\frac{0}{0}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we multiply the numerator and denominator by the conjugate of the numerator.
-|-
-|Hence, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}} & = & \displaystyle{\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}\frac{(\sqrt{x+5}+3)}{(\sqrt{x+5}+3)}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow 4} \frac{(x+5)-9}{(x-4)(\sqrt{x+5}+3)}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow 4} \frac{x-4}{(x-4)(\sqrt{x+5}+3)}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow 4} \frac{1}{\sqrt{x+5}+3}}\\
-&&\\
-& = & \displaystyle{ \frac{1}{\sqrt{9}+3}}\\
-&&\\
-& = & \displaystyle{\frac{1}{6}.}
-\end{array}</math>
-|-
-|
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We proceed using L'Hôpital's Rule. So, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow 0} \frac{\sin^2 (x)}{3x}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 0}\frac{2\sin(x)\cos(x)}{3}.}
-\end{array}</math>
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we plug in &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; to get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow 0} \frac{\sin^2 (x)}{3x}} & = & \displaystyle{\frac{2\sin(0)\cos(0)}{3}}\\
-&&\\
-& = & \displaystyle{\frac{2(0)(1)}{3}}\\
-&&\\
-& = & \displaystyle{0.}
-\end{array}</math>
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2(1+\frac{2}{x^2})}}{2x-1}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{|x|\sqrt{1+\frac{2}{x^2}}}{2x-1}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-x\sqrt{1+\frac{2}{x^2}}}{x(2-\frac{1}{x})}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-1\sqrt{1+\frac{2}{x^2}}}{(2-\frac{1}{x})}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-1\sqrt{1+\frac{2}{x^2}}}{(2-\frac{1}{x})} }\\
-&&\\
-& = & \displaystyle{\frac{-\sqrt{1+0}}{(2-0)}}\\
-&&\\
-& = & \displaystyle{-\frac{1}{2}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>\frac{1}{6}</math>
-|-
-|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>0</math>
-|-
-|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp;<math>-\frac{1}{2}</math>
-|}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Final 2, Problem 1"

Latest revision as of 09:05, 1 December 2017

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